<span> binomdist with n = 3, p = 0.82, q = 1-0.82 = 0.18
P[≥2] = P[2] + P[3] = 3c2 *0.82^2*0.18 + 0.82^3 ≈ 91%
hope it helps
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The probability he will select a hockey card is 1/4, and then the probability that he selects a baseball card without replacement is 1/6.
The answer is 8.031 per person
